If the sum of all solutions of the equation 8 x(6+x)(6-x) - 12 = 1 in [0,] is k, then k is…

Question

If the sum of all solutions of the equation $8\cos x\cos\left(\dfrac\pi6+x\right)\cos\left(\dfrac\pi6-x\right) - \dfrac12 = 1$ in $[0,\pi]$ is $k\pi$, then $k$ is equal to:

Accepted Answer

Using $\cos\left(\tfrac\pi6+x\right)\cos\left(\tfrac\pi6-x\right) = \cos^2\tfrac\pi6-\sin^2x = \tfrac34-\sin^2x = \tfrac34-(1-\cos^2x) = \cos^2x-\tfrac14$:

$$8\cos x\left(\cos^2x-\tfrac14\right) - \tfrac12 = 1 \implies 8\cos^3x - 2\cos x - \tfrac12=1 \implies 2(4\cos^3x-3\cos x)=1$$

$$2\cos3x = 1 \implies \cos3x = \frac12 \implies 3x = 2n\pi\pm\frac\pi3 \implies x = \frac{2n\pi}3\pm\frac\pi9$$

Within $[0,\pi]$: $x = \dfrac\pi9,\ \dfrac{2\pi}3+\dfrac\pi9,\ \dfrac{2\pi}3-\dfrac\pi9$.

$$\text{Sum} = \frac\pi9+\frac{2\pi}3+\frac\pi9+\frac{2\pi}3-\frac\pi9 = \frac\pi9+\frac{4\pi}3 = \frac{\pi+12\pi}9 = \frac{13\pi}9$$

Solutions · 1